Does Σ 1/(3n - 2) converge Step by Step

Step by step

The question

Question: decide if the series converges: the sum of 1/(3n − 2) for n from 1 to infinity.

Σ_{n = 1}^∞13n − 2 =

1 For big n

For big n, only the highest powers matter, top and bottom.

a_n = 13n − 2 ≈ 13n

2 The known series

We know the sum of 1/n: a p-series with p = 1.

p is 1, so it diverges.

A number in front, 1/3, does not change that.

Σ_{n = 1}^∞1n: p = 1, diverges

3 Compare

The bottom has a minus term: it is smaller than 3n.

3n − 2 < 3n

A smaller bottom makes a bigger fraction: every term is bigger than 1/(3n).

a_n > 13n

The direct comparison test: our terms are bigger than terms whose sum grows for ever.

So ours grow for ever too.

bigger than a diverging series: diverges

4 Decide

So our series diverges.

= diverges

Careful!

Smaller than a series that diverges tells us nothing.

We need bigger than a diverging one, or the limit test.

Careful: not a_n < 1n, so it converges ✗

5 Check

Let's check with a calculator: 100, 1000 and 10000 terms add up to about 2.58, 3.35 and 4.11.

It keeps growing.

Right!

S₁₀₀ ≈ 2.5785, S₁₀₀₀ ≈ 3.3465, S₁₀₀₀₀ ≈ 4.1141

Let's look back.

First: for big n.

Then: the known series.

Then: compare.

Then: decide.

Last: check.

Answer